I assume working with stabilizer codes where the stabilizer group is denoted $S$ and the code space is $C(S)$.

The minimal requirement for a logical operation $M$ is to have:

$$\forall |\psi\rangle \in C(S), M |\psi \rangle \in C(S)$$

Indeed, in such case I apply a unitary operation that will make sure that any state in the code space remain in it. I am wondering if there is a nice general characterization of such operations.

I take an example. The normalizer of $S$ in $U(2^n)$ (the unitary matrices acting on $n$ qubits) is composed of valid logical operations because any matrix $M$ in it will verify:

$$\forall g \in S, \exists g' \in S | M g = g' M$$ It implies: $$\forall |\psi \rangle \in C(S), \forall g \in S \ \exists g' | \ g(M | \psi \rangle) = g M |\psi \rangle = M g'|\psi \rangle=M |\psi \rangle$$

This last equation shows that $M |\psi \rangle$ will be stabilized by any $g \in S$ and will thus remain in the code space. So, any element in the normalizer of $S$ in $U(2^n)$ is a valid logical operation, but there might exist logical operation outside of this set (this is not an equivalence, I did not characterized the logical operations)

In conclusion: is there a general characterization of logical operations ? Like they live in some well defined group ?


3 Answers 3


For a stabilizer code $S$ (i.e. an abelian subgroup of the $n$-qubit Pauli group $G$), the normalizer of $S$ in $G$ is denoted by $N_G(S)$. Following Preskill's notes, the logical Pauli operators are the elements in $N_G(S)\setminus S$, where "$\setminus$" is the subtraction of sets. It follows that the set of logical operators $N_G(S)\setminus S$ cannot form a group, since the identity operator is always in $S$.

It is also worth remarking that for the Pauli group, the normalizer of $S$ in $G$ is equal to the centralizer of $S$ in $G$, which is why one might see the centralizer condition instead of the normalizer condition when reading about stabilizer codes.

  • 1
    $\begingroup$ Thank you for your answer. I am not sure to understand why it is the appropriate set. For instance a transversal cNOT is a logical gate for Steane and it is not an n-Pauli matrix. Thus it is not in the normalizer of S in G. And even for single qubit logical gate I dont see why they would necessarily be in this set. $\endgroup$ Commented Sep 17, 2021 at 17:31
  • 1
    $\begingroup$ A transversal Hadamard for instance cannot be in the normalizer of S in the Pauli group as it is not a Pauli matrix. It could be in the normalizer of S in the unitary matrices set however. But then even if we take this bigger group I dont see in principle why element outside of this bigger group could not still be logical operators $\endgroup$ Commented Sep 17, 2021 at 17:37
  • $\begingroup$ How do you know that these transversal gates you mention aren't in the $n$-Pauli group? Just because $CNOT$ and $H$ aren't in the Pauli group doesn't mean that their representatives on the encoded space aren't. $\endgroup$
    – Condo
    Commented Sep 17, 2021 at 17:50
  • $\begingroup$ Because a logical hadamard can be implemented by applying an hadamard on each physical qubit composing the logical qubit for Steane code for instance. $\endgroup$ Commented Sep 17, 2021 at 17:51
  • $\begingroup$ hmmm I see your point. I was simply following the textbook definition of a "logical operator", which perhaps only refers to employing elements of $G$ on $S$, which is sufficient for error correction. $\endgroup$
    – Condo
    Commented Sep 17, 2021 at 18:03

TL;DR: Logical operations live in the set of operators $U_L\oplus U_N$ where $U_L$ acts on the code subspace and $U_N$ acts on its orthogonal complement. For a fixed logical operation $M$ there are many operators on the Hilbert space of the underlying $n$ physical qubits that realize $M$ within the code subspace. There is a subgroup $\mathcal{A}$ of the centralizer $C_{U(2^n)}(S)$ of $S$ in $U(2^n)$ which is isomorphic to the set of logical operations. Finally, logical operations do have realizations that live outside $\mathcal{A}$ and even outside of the normalizer $N_{U(2^n)}(S)$ of $S$ in $U(2^n)$. Below, we exhibit an example of such a realization for the three qubit repetition code.

Denote with $\mathcal{L}(S)$ the set of logical operations in an $[\![n, k]\!]$ code with the stabilizer group $S$. In other words, $M\in\mathcal{L}(S)$ if and only if

$$ \forall |\psi\rangle \in C(S) \,\, M|\psi \rangle \in C(S) $$

as defined in the question. Recall that $n$ is the number of physical qubits and $k$ the number of logical qubits.

Characterization of operators preserving code subspace

Denote with $\mathcal{H}$ the finite-dimensional Hilbert space of the $n$ physical qubits of which $C(S)$ is a subspace. Denote with $C(S)^\perp$ the orthogonal complement of $C(S)$. By definition, we have

$$ \mathcal{H}=C(S)\oplus C(S)^\perp\tag1. $$

Consider a unitary operator $U$ on $\mathcal{H}$ of the form

$$ U = U_L\oplus U_N\tag2 $$

where $U_L$ is a unitary operator on $C(S)$ and $U_N$ a unitary operator on $C(S)^\perp$. It is clear that $U_L$ is the logical operation corresponding to the action of $U$ on the physical qubits. Conversely, if $U_L$ is a logical operation on $C(S)$ then for any unitary $U_N$ acting on $C(S)^\perp$ the operator $U=U_L\oplus U_N$ acting on $\mathcal{H}$ applies the logical operation $U_L$ in the code subspace $C(S)$.

Thus, logical operations correspond to operators that can be written in the form $(2)$.

Group structure

Operators of the form $(2)$ live in the group $U(2^k)\oplus U(2^n-2^k)$. However, the correspondence between logical operations on $C(S)$ and operators of the form $(2)$ is not one-to-one as operators $U_L\oplus U_N$ and $U_L\oplus U_N'$ with $U_N\ne U_N'$ correspond to the same logical operation. Therefore, the logical operations live in

$$ \left(U(2^k) \oplus U(2^n-2^k)\right)\big/_{U(2^n-2^k)} \cong U(2^k)\tag3 $$

where we quotient out the irrelevant $U_N$. Disregarding the global phase, we see that the logical operations live in the projective unitary group on $k$ qubits

$$ \mathcal{L}(S) \cong PU(2^k).\tag4 $$

This is expected as the logical operations are quantum gates acting on $k$ qubits. Note that the physical and logical operations on a given number of qubits must possess the same group structure if logical qubits are to be used as a noise-free replacement for physical qubits. This is confirmed in $(4)$.

Matrix representation

Any basis of $C(S)$ can be extended to a basis of $\mathcal{H}$. Therefore, there exists a basis of $\mathcal{H}$

$$ |0\dots 0_L\rangle, \dots, |1\dots 1_L\rangle, |N_1\rangle, \dots, |N_{2^n-2^k}\rangle\tag5 $$

where the $2^k$ states $|0\dots 0_L\rangle$ through $|1\dots 1_L\rangle$ make up the logical computational basis and the $2^n-2^k$ states $|N_1\rangle$ through $|N_{2^n-2^k}\rangle$ help protect the logical subspace from noise. In this basis, the matrix of operators of the form $(2)$ has the following block-diagonal structure

$$ U=\begin{bmatrix} \begin{array}{c|c} U_L&\\ \hline &U_N \end{array} \end{bmatrix} $$

where empty blocks are filled with zeros which correspond to the fact that $U$ maps states in $C(S)$ to states in $C(S)$ and states in $C(S)^\perp$ to states in $C(S)^\perp$.

Centralizer and normalizer of $S$

Define the set of unitary operators on $\mathcal{H}$ that act as identity on $C(S)^\perp$

$$ \mathcal{A} = \{U\,|\,U = M\oplus I_{2^n-2^k}\}\tag6 $$

where $M$ acts on $C(S)$ and the identity on $C(S)^\perp$. This set is a subset of the centralizer $C_{U(2^n)}(S)$ of $S$ in $U(2^n)$.

To see this inclusion, note that by definition of $S$, for any $|\phi\rangle \in C(S)$ and any $g\in S$ we have $g|\phi\rangle = |\phi\rangle$. Since the equality holds for all $|\phi\rangle\in C(S)$ we can write $g$ in the basis $(5)$ as

$$ g = I_{2^k}\oplus g' $$

for some $g'\in U(2^n-2^k)$. Comparing to $(6)$, we see that $g$ and $U$ commute for all $g\in S$ and all $U\in \mathcal{A}$. Thus, $\mathcal{A} \subset C_{U(2^n)}(S)$.

It is not hard to see that $\mathcal{L}(S)$ is isomorphic to $\mathcal{A}$. The isomorphism $f: \mathcal{L}(S) \to \mathcal{A}$ is defined as

$$ f(M) = M\oplus I_{2^n-2^k} $$

where $M$ acts on $C(S)$ and the identity on $C(S)^\perp$.

Note that there are elements in $C_{U(2^n)}(S)$ that do not belong to $\mathcal{A}$. For example, $U=M\oplus g'$ for any $M\in\mathcal{L}(S)$ and any $g'\ne I$ such that $I_{2^k}\oplus g'\in S$. These are additional realizations of $M$ whose action on the irrelevant $C(S)^\perp$ is non-identity, in contrast to $M\oplus I\in\mathcal{A}$. In fact, there are operators outside of the normalizer $N_{U(2^n)}(S)$ that realize a given logical operation on $C(S)$ as the following example shows.


Consider the three-qubit repetition code with logical states $|0_L\rangle=|000\rangle$ and $|1_L\rangle=|111\rangle$ and stabilizer group

$$ S = \{III, IZZ, ZIZ, ZZI\}. $$

Note that all elements of $S$ are diagonal. Therefore, if a unitary $U$ is such that $UgU^\dagger$ is not diagonal then $U$ does not belong to the normalizer $N_{U(2^n)}(S)$ of $S$ in $U(2^n)$.

Consider the operator $W$ defined as

$$ W=\begin{bmatrix} & & & & & & & 1 \\ & 1 & & & & & & \\ & & 1 & & & & & \\ & & & 1 & & & & \\ & & & & \frac{1}{\sqrt2} & \frac{1}{\sqrt2} & & \\ & & & & \frac{1}{\sqrt2} & -\frac{1}{\sqrt2} & & \\ & & & & & & 1 & \\ 1 & & & & & & & \\ \end{bmatrix}. $$

Clearly $W$ is unitary and realizes logical $X$ on the code subspace $C(S)=\mathrm{span}(|000\rangle, |111\rangle)$. Note that the restriction $W^\perp$ of $W$ to $C(S)^\perp$ can be written as

$$ W^\perp = I_3 \oplus H \oplus I_1 $$

and the restriction $IZZ^\perp$ of $IZZ$ to $C(S)^\perp$ can be written as

$$ \begin{align} IZZ^\perp &= \mathrm{diag}(-1, -1, 1, 1, -1, -1) \\ &= (-I_1) \oplus (-Z) \oplus Z \oplus (-I_1). \end{align} $$


$$ W^\perp\circ IZZ^\perp \circ {W^\perp}^\dagger= (-I_1) \oplus (-Z) \oplus X \oplus (-I_1) $$


$$ \begin{align} W\circ IZZ \circ W^\dagger &= Z \oplus (-Z) \oplus X \oplus (-Z) \\ &=\begin{bmatrix} 1 & & & & & & & \\ & -1 & & & & & & \\ & & -1 & & & & & \\ & & & 1 & & & & \\ & & & & 0 & 1 & & \\ & & & & 1 & 0 & & \\ & & & & & & -1 & \\ & & & & & & & 1 \\ \end{bmatrix} \end{align} $$

and thus $W\notin N_{U(2^n)}(S)$.


From $(4)$, we see that $\mathcal{L}(S)$ is infinite.

On the other hand, $N_G(S)\subset G$ and the $n$-qubit Pauli group $G$ are finite. Similarly, $S$ is finite, so the automorphism group $\mathrm{Aut}(S)$ of $S$ and its subset the outer automorphism group $\mathrm{Out}(S)$ are both finite. See for example this question.

We conclude that

$$ \begin{align} \mathcal{L}(S)&\ne \mathrm{Aut}(S)\\ \mathcal{L}(S)&\ne \mathrm{Out}(S)\\ \mathcal{L}(S)&\ne N_G(S)-S \end{align} $$

which rules out the characterizations of $\mathcal{L}(S)$ proposed in the other two answers.

Normalizer in $G$

The elements of $N_G(S)-S$ are indeed logical operations. However, since the set is a subset of the $n$-qubit Pauli group it only contains a small subset of logical operations. Namely, those that take the form of $n$-qubit Pauli operators.

Automorphism group

The idea that the automorphism group of $S$ gives the structure of the group of logical operators is wrong because most states in $C(S)$ are not stabilizer states$^1$. Consequently, transformations of the stabilizer group are insufficient to describe the logical operators acting on $C(S)$.

$^1$ Stabilizer states are defined as states stabilized by $2^n$ Pauli operators. Most states in $C(S)$ are stabilized by exactly $2^k < 2^n$ Pauli operators.
  • $\begingroup$ Comments are not for extended discussion; this conversation has been moved to chat. $\endgroup$
    – glS
    Commented Sep 27, 2021 at 23:30
  • 2
    $\begingroup$ Thank you a lot for this very detailed answer. I have now very full weeks so I cannot read it in details now. But when I do I come back to you and I may award you the bounty then. $\endgroup$ Commented Sep 28, 2021 at 9:44

In the context of stabilizer QECCs, the interesting/useful logical operations live in the set of valid fault-tolerant encoded operations. The automorphism group of $S$, $\text{Aut}(S)$, determines this set of valid fault-tolerant encoded operations.

$S$ is necessarily Abelian, and Abelian groups have no non-trivial inner-automorphisms. So more precisely, the relevant group is the group of outer-automorphisms of $S$, $\text{Out}(S)$.

Gottesman's Thesis is a great primer on this subject.

Edit: The conversation spawned by the other answers suggests some confusion about the relationship between $U(2)$, $SU(2)$, $PU(2)$, and $\mathbb{CP}^1$. So I'm adding some clarifying commentary below.

$\mathbb{CP}^1$ is a great space to represent qubit states, but $PU(2)=SO(3)$ does not represent gate operations on qubits, and doesn't act on $\mathbb{CP}^1$. Niether of these groups has a two-dimensional representation, and the Bloch sphere is represented over $\mathbb{C}^2$.

The natural action of $PU(2)=SO(3)$ is adjoint action on $SU(2)$, i.e. action on the tangent space of $SU(2)$. To see this, set $G \equiv SU(2)$, with automorphisms of $G$ by conguation given by $$\Psi_g:G\rightarrow Aut(G), \;\; \Psi_g(h)=ghg^{-1}, \;\; g,h \in SU(2).$$ The adjoint representation, $Ad(g)$, is given by the differential of $\Psi$ at the identity ($I_2$): $$Ad:G\rightarrow Aut(T_{I_2}G), \;\; Ad(g)=(d\Psi_g)_{I_2}: T_{I_2}G \rightarrow T_{I_2}G.$$

All this fits together nicely, the tangent space of $SU(2)$ is a Euclidean space (this is a fundamental property of both group manifolds and Riemann manifolds), and this particular Euclidean space is $\mathbb{R}^3$. The adjoint representation of $SU(2)$, which is $PU(2)=SO(3)$ acts on $\mathbb{R}^3$ to describe homomorphisms of the tangent space of $SU(2)$.

Since $Ad$ is the differential of a conjugation map, its obvious that phase is in the kernel of $Ad$. That doesn't mean that $PU(2)=SO(3)$ acts on $\mathbb{CP}^1$ though. In fact any possible action on a qubit of any possible gate is by conjugation, so phase is in the kernel of all possible relevant gate representations.

Regardless, if one really wants to look at actions on $\mathbb{CP}^1$, no need to go any further than $\mathbb{CP}^1$. Since $U(1)$ is a normal subgroup of $U(2)$, $\mathbb{CP}^1=U(2)/U(1)$ is a quotient group and has the structure of both a differentiable manifold and Lie group. So you can look at its action on itself by left/right multiplication, automorphisms, etc., just like you would $U(2)$ or any other Lie group.

There are other ways to look at this too, but they all lead to the same conclusion. For example, you could talk about it in more geometric language, where $SU(2)$ (covering space) is a $\mathbb{Z}_2$ bundle over $PU(2)$ (base space), and get the same answers.

  • $\begingroup$ Comments are not for extended discussion; this conversation has been moved to chat. $\endgroup$
    – glS
    Commented Sep 28, 2021 at 8:14

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.